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What is the multiplicative inverse of 4 plus 3i?
Use the rules of division for complex numbers. Just divide 1 / (4 + 3i). This requires multiplying numerator and denominator of this fraction by (4 - 3i), to get a real number in the denominator.
How do you solve complex fraction?
this is a very good question. lets solve (2+3i)/(4-2i). we want to make 4-2i real by multiplying it by the conjugate, or 4+2i (4-2i)(4+2i)=16-8i+8i+4=20, now we have (2+3i)/20 0r 1/10 + 3i/20 notice that -2i times 2i = -4i^2 =-4 times -1 = 4
If the sum of 2 numbers is 4 and the product of these numbers is 13 what are these 2 numbers?
The question has no answer in real numbers. The solution, in complex numbers, are 2+3i and 2-3i where i is the imaginary square root of -1.
What multiplies to 105 and adds to 8?
113
What is the factorization of T power 4 - 81?
-77
What is the complex conjugate of 3i?
0+3i has a complex conjugate of 0-3i thus when you multiply them together (0+3i)(0-3i)= 0-9i2 i2= -1 0--9 = 0+9 =9 conjugates are used to eliminate the imaginary parts
What is -4-3i divided by 4 plus i?
To divide by a complex number, write it as a fraction and then multiply the numerator and denominator by the complex conjugate of the denominator - this is formed by changing the sign of the imaginary bit of the number; when a complex number (a + bi) is multiplied by its complex conjugate the result is the real number a² + b² which can be divided into the complex number of the numerator: (-4 - 3i) ÷ (4 + i) = (-4 - 3i)/(4 + i) = ( (-4 - 3i)×(4 - i) ) / ( (4 + i)×(4 - i) ) = (-16 + 4i - 12i + 3i²) / (4² + 1²) = (-16 - 8i - 3) / (16 + 1) = (-19 - 8i)/17
What is the multiplicative inverse of 4 plus 3i?
Use the rules of division for complex numbers. Just divide 1 / (4 + 3i). This requires multiplying numerator and denominator of this fraction by (4 - 3i), to get a real number in the denominator.
Complex conjugate of 5i-4?
To get the complex conjugate, change the sign in front of the imaginary part. Thus, the complex conjugate of -4 + 5i is -4 - 5i.
How do you solve complex fraction?
this is a very good question. lets solve (2+3i)/(4-2i). we want to make 4-2i real by multiplying it by the conjugate, or 4+2i (4-2i)(4+2i)=16-8i+8i+4=20, now we have (2+3i)/20 0r 1/10 + 3i/20 notice that -2i times 2i = -4i^2 =-4 times -1 = 4
Which polynomial has rational coefficients a leading leading coefficient of 1 and the zeros at 2-3i and 4?
There cannot be such a polynomial. If a polynomial has rational coefficients, then any complex roots must come in conjugate pairs. In this case the conjugate for 2-3i is not a root. Consequently, either (a) the function is not a polynomial, or (b) it does not have rational coefficients, or (c) 2 - 3i is not a root (nor any other complex number), or (d) there are other roots that have not been mentioned. In the last case, the polynomial could have any number of additional (unlisted) roots and is therefore indeterminate.
Find the quotient of these complex numbers (6-i)/(4 3i)?
4/3 -2i
How do you do the math problem (4-3i)(5 2i)?
To multiply complex numbers you can use the same FOIL rule that you use for multiplying binomials (First, Inside, Outside, Last).(4 - 3i)(5 + 2i) = (4)(5) +(4)(2i) - (3i)(5) - (3i)(2i) = 20 + 8i-15i - 6(i)^2= 20 -7i - 6(-1) = 20 + 6 -7i = 26 -7i.
Obtain fifth roots of 4 3i?
Obtain fifth roots of 4+3i Obtain fifth roots of 4+3i
Let A equals 1 plus 2i and let C equals 5 plus 5i what is A-C?
-4-3i
Which one is greater number 2 plus 3i or 3 plus 2i?
the magnitude of a complex number is the square root of the sum of the squares of the real and imaginary parts. so......... (4+9)^.5 = ~3.6 and also (9+4)^.5 = ~3.6 so they are both the same.
If the sum of 2 numbers is 4 and the product of these numbers is 13 what are these 2 numbers?
The question has no answer in real numbers. The solution, in complex numbers, are 2+3i and 2-3i where i is the imaginary square root of -1.
